4
$\begingroup$

Suppose $\mathbb{P}_0$, $\mathbb{P}_1$ and $\mathbb{P}_2$ are separative posets such that $\mathbb{P}_2$ projects into $\mathbb{P}_1$ and $\mathbb{P}_1$ projects into $\mathbb{P}_0$, i.e. there are projections of the form $\pi_{2,1}:\mathbb{P}_2\rightarrow\mathbb{P}_1$ and $\pi_{1,0}:\mathbb{P}_1\rightarrow\mathbb{P}_0$. For $i\in\{0,1,2\}$, let $\mathbb{B}_i$ be the boolean completion of $\mathbb{P}_i$. Using the composition $\pi_{2,0}:=\pi_{1,0}\circ\pi_{2,1}$, one can define a complete embedding $i$ from $\mathbb{B}_0$ to $\mathbb{B}_2$, by setting $$i(p_0)=\bigvee\{p_2\in\mathbb{P}_2\mid\pi_{2,0}(p_2)\leq p_0\},$$ for all $p_0\in\mathbb{P}_0$. I wonder whether this complete embedding admits some kind of factorization. More specifically, suppose $j:\mathbb{B}_0\rightarrow\mathbb{B}_1$ is an arbitrary complete embedding.

Question. Can we find a complete embedding $k:\mathbb{B}_1\rightarrow\mathbb{B}_2$ such that $k\circ j=i$?

I see the question is very general and the answer could be no, since there are no information on $j$. So I would like to see if the question has a positive answer under the following additional assumptions.

Let $\dot g_0$ be a $\mathbb{P}_1$-name such that $\Vdash_{\mathbb{P}_1}``\dot g_0$ is $\mathbb{P}_0$-generic$"$, let $G_2$ be $\mathbb{P}_2$-generic, and define $g_0=(\dot g_0)_{\pi_{2,1}``G_2}$ (i.e. $g_0$ is the interpretation of $\dot g_0$ by the $\mathbb{P}_1$-generic generated by $\pi_{2,1}``G_2$). Suppose there is a $\mathbb{P}_1$-generic $g_1\in V[G_2]$ such that $\pi_{1,0}``g_1\subseteq g_0$ (in other words, $g_1$ is $\mathbb{P}_1/g_0$-generic over $V[g_0]$). Define $$j(p_0)=\bigvee\{p_1\in\mathbb{P}_1\mid p_1\Vdash p_0\in\dot g_0\},$$ for all $p_0\in\mathbb{P}_0$. So now $j$ is no longer an arbitrary complete embedding, but a very specific one. The question is the same as above.

Question. Is there a complete embedding $k:\mathbb{B}_1\rightarrow\mathbb{B}_2$ such that $k\circ j=i$?

A candidate that I have in mind is $k(p_1)=\bigvee\{p_2\in\mathbb{P}_2\mid p_2\Vdash p_1\in \dot g_1\}$ for $p_1\in\mathbb{P}_1$, where $\dot g_1$ is a $\mathbb{P}_2$-name for $g_1$ (recall that $g_1\in V[G_2]$). However, I can't prove that the diagram commutes (i.e. I'm not sure $k\circ j=i$).

$\endgroup$

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.